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Chapter 6 DC Machines EET103/4 Introduction • An electrical machine is link between an electrical system and a mechanical system. • Conversion from mechanical to electrical: generator • Conversion from electrical to mechanical: motor Introduction Machines are called • AC machines (generators or motors) if the electrical system is AC. • DC machines (generators or motors) if the electrical system is DC. DC machines can be divide by: a) DC motor b) DC Generator DC Machines DC Motor DC Generator DC Machines Construction cutaway view of a dc machine DC Machines Construction cutaway view of a dc machine DC Machines Construction Rotor of a dc machine DC Machines Construction Stator of a dc machine DC Machines Fundamentals • Stator: is the stationary part of the machine. The stator carries a field winding that is used to produce the required magnetic field by DC excitation. • Rotor (Armature): is the rotating part of the machine. The rotor carries a distributed winding, and is the winding where the e.m.f. is induced. • Field winding: Is wound on the stator poles to produce magnetic field (flux) in the air gap. • Armature winding: Is composed of coils placed in the armature slots. • Commutator: Is composed of copper bars, insulated from each other. The armature winding is connected to the commutator. • Brush: Is placed against the commutator surface. Brush is used to connect the armature winding to external circuit through commutator DC Machines Fundamentals In DC machines, conversion of energy from electrical to mechanical form or vice versa results from the following two electromagnetic phenomena 1.When a conductor moves in a magnetic field, voltage is induced in the conductor. 2. When a current carrying conductor is placed in magnetic field, the conductor experiences a mechanical forces. DC Machines Fundamentals Generator action: An e.m.f. (voltage) is induced in a conductor if it moves through a magnetic field. Motor action: A force is induced in a conductor that has a current going through it and placed in a magnetic field •Any DC machine can act either as a generator or as a motor. DC Machines Equivalent Circuit The equivalent circuit of DC machines has two components: Armature circuit: • It can be represented by a voltage source and a resistance connected in series (the armature resistance). The armature winding has a resistance, RA. The field circuit: • It is represented by a winding that generates the magnetic field and a resistance connected in series. The field winding has resistance RF. DC Motor Basic Operation of DC Motor • In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field. • The rotor is supplied by dc current through the brushes, commutator and coils. • The interaction of the magnetic field and rotor current generates a force that drives the motor. Basic Operation of DC Motor • The magnetic field lines enter into the rotor from the north pole (N) and exit toward the south pole (S) • The poles generate a magnetic field that is perpendicular to the current carrying conductors • The interaction between the field and the current produces a Lorentz force • The force is perpendicular to both the magnetic field and conductor Basic Operation of DC Motor • The generated force turns the rotor until the coil reaches the neutral point between the poles. • At this point, the magnetic field becomes practically zero together with the force. • However, inertia drives the motor beyond the neutral zone where the direction of the magnetic field reverses. • To avoid the reversal of the force direction, the commutator changes the current direction, which maintains the counter clockwise rotation. Basic Operation of DC Motor • Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2 • Therefore, current enters the coil end at slot ‘a’ and exits from slot ‘b’ during this stage • After passing the neutral zone, the current enters segment 2 and exits from segment 1, • This reverses the current direction through the rotor coil, when the coil passes the neutral zone • The result of this current reversal is the maintenance of the rotation Basic Operation of DC Motor Classification of DC Motor 1. Separately Excited DC Motor • Field and armature windings are either connected separate. 2. Shunt DC Motor • Field and armature windings are either connected in parallel. 3. Series DC Motor • Field and armature windings are connected in series. 4. Compound DC Motor • Has both shunt and series field so it combines features of series and shunt motors. Important terms • • • • • • • • • • VT – supply voltage EA – internal generated voltage/back e.m.f. RA – armature resistance RF – field/shunt resistance RS – series resistance IL – load current IF – field current IA – armature current IL – load current n – speed Generated or back e.m.f. of DC Motor • General form of back e.m.f., ZN P EA 60 A Φ = flux/pole (Weber) Z = total number of armature conductors = number of slots x number of conductor/slot P = number of poles A = number of parallel paths in armature [A = 2 (for wave winding), A = P (for lap winding)] N = armature rotation (rpm) EA = back e.m.f. Torque Equation of a DC Motor • The armature torque of a DC motor is given by ZI A P Ta ( Newton meter) 2 A Φ = flux/pole (Weber) Z = total number of armature conductors = number of slots x number of conductor/slot P = number of poles A = number of parallel paths in armature IA = armature current Ta = armature torque Equivalent Circuit of DC Motor Separately Excited DC Motor VF IF RF IL IA VT E A I A RA Shunt DC Motor VT IF RF IL I A IF VT E A I A RA Series DC Motor VT E A I A ( RA RS ) I A IS IL Compound DC Motor VT IF RF I A IL IF VT E A I A ( RA RS ) Speed of a DC Motor • For shunt motor n2 E A 2 1 n1 E A1 2 • For series motor n2 E A2 If 2 1 , then n1 E A1 n2 E A 2 1 E A 2 I A1 n1 E A1 2 E A1 I A 2 Example 1 A 250 V, DC shunt motor takes a line current of 20 A. Resistance of shunt field winding is 200 Ω and resistance of the armature is 0.3 Ω. Find the armature current, IA and the back e.m.f., EA. Solution Given quantities: • Terminal voltage, VT = 250 V • Field resistance, RF = 200 Ω • Armature resistance, RA = 0.3 Ω • Line current, IL = 20 A Figure 1 Solution (cont..) IL IA IF the field current, V T 250V IF 1.25A RF 200 the armature current, I A IL IF 20A 1.25A 18.75A VT = EA + IARA the back e.m.f., EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V Example 2 A 50hp, 250 V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding. Example 2 (cont..) a) Find the speed of this motor when its input current is 100 A. b) Find the speed of this motor when its input current is 200 A. c) Find the speed of this motor when its input current is 300 A. Solution Given quantities: • Terminal voltage, VT = 250 V • Field resistance, RF = 50 Ω • Armature resistance, RA = 0.06 Ω • Initial speed, n1 = 1200 r/min Figure 2 Solution (cont..) (a) When the input current is 100A, the armature current in the motor is VT I A IL IF IL RF 250V 100 A 50 100 A 5 A 95 A Therefore, EA at the load will be E A VT I A RA 250V (95 A)(0.06 ) 250 V 5.7 V 244.3V Solution (cont..) • The resulting speed of this motor is n2 E A2 n1 E A1 E A2 n2 n1 E A1 244.3V 1200 r / min 250V 1173 r / min Solution (cont..) (b) When the input current is 200A, the armature current in the motor is VT I A IL IF IL RF 250V 200 A 50 200 A 5 A 195 A Therefore, EA at the load will be E A VT I A RA 250V (195 A)(0.06 ) 250 V 11.7 V 238.3 V Solution (cont..) • The resulting speed of this motor is n2 E A2 n1 E A1 E A2 n2 n1 E A1 238.3V 1200 r / min 250V 1144 r / min Solution (cont..) (c) When the input current is 300A, the armature current in the motor is VT I A IL IF IL RF 250V 300 A 50 300 A 5 A 295 A Therefore, EA at the load will be E A VT I A RA 250V (295 A)(0.06 ) 250 V 17.7 V 232.3 V Solution (cont..) • The resulting speed of this motor is n2 E A2 n1 E A1 E A2 n2 n1 E A1 232.3 V 1200 r / min 250 V 1115 r / min Example 3 The motor in Example 2 is now connected in separately excited circuit as shown in Figure 3. The motor is initially running at speed, n = 1103 r/min with VA = 250 V and IA = 120 A, while supplying a constant-torque load. If VA is reduced to 200 V, determine i). the internal generated voltage, EA ii). the final speed of this motor, n2 Example 3 (cont..) Figure 3 Solution Given quantities • • • • Initial line current, IL = IA = 120 A Initial armature voltage, VA = 250 V Armature resistance, RA = 0.06 Ω Initial speed, n1 = 1103 r/min Solution (cont..) i) The internal generated voltage EA = VT - IARA = 250 V – (120 A)(0.06 Ω) = 250 V – 7.2 V = 242.8 V Solution (cont..) ii) Use KVL to find EA2 EA2 = VT - IA2RA Since the torque is constant ant he flux is constant, IA is constant. This yields a voltage of EA2 = 200 V – (120 A)(0.06 Ω) = 200 V – 7.2 V = 192.8 V Solution (cont..) • The final speed of this motor n2 E A2 n1 E A1 E A2 n2 n1 E A1 192.8 V 1103 r / min 242.8 V 876 r / min Example 4 A DC series motor is running with a speed of 800 r/min while taking a current of 20 A from the supply. If the load is changed such that the current drawn by the motor is increased to 50 A, calculate the speed of the motor on new load. The armature and series field winding resistances are 0.2 Ω and 0.3 Ω respectively. Assume the flux produced is proportional to the current. Assume supply voltage as 250 V. Solution Given quantities • Supply voltage, VT = 250 V • Armature resistance, RA = 0.2 Ω • Series resistance, RS = 0.3 Ω • Initial speed, n1 = 800 r/min • Initial armature current, Ia1 = IL1 = 20 A Figure 4 Solution (cont..) For initial load, the armature current, Ia1 = 20 A and the speed n1 = 800 r/min V = EA1 + Ia1 (RA + RS) The back e.m.f. at initial speed EA1 = V - Ia1 (RA + RS) = 250 – 20(0.2 + 0.3) = 240 V Solution (cont..) When the armature current increased, Ia2 = 50 A, the back emf n2 E A 2 1 EA2 = V – Ia2 (RA + RS) n1 E A1 2 = 250 – 50(0.2 + 0.3) = 225 V n2 E A 2 I1 n1 The speed of the motor on new load E A1 I2 E A 2 I1 n2 n1 E A1 I 2 225 20 800 240 50 300 r / min DC Generator Generating of an AC Voltage • The voltage generated in any dc generator inherently alternating and only becomes dc after it has been rectified by the commutator Generation of an AC Voltage Armature windings • The armature windings are usually formerwound. This are first wound in the form of flat rectangular coils and are then puller. • Various conductors of the coils are insulated each other. The conductors are placed in the armature slots which are lined with tough insulating material. • This slot insulation is folded over above the armature conductors placed in the slot and is secured in place by special hard wooden or fiber wedges. Lap and wave Windings There are two types of windings mostly employed: • Lap winding • Wave winding The difference between the two is merely due to the different arrangement of the end connection at the front or commutator end of armature. Generated or back e.m.f. of DC Generator • General form of generated e.m.f., ZN P E 60 A Φ = flux/pole (Weber) Z = total number of armature conductors = number of slots x number of conductor/slot P = number of poles A = number of parallel paths in armature [A = 2 (for wave winding), A = P (for lap winding)] N = armature rotation (rpm) E = e.m.f. induced in any parallel path in armature Classification of DC Generator 1. Separately Excited DC Generator • Field and armature windings are either connected separate. 2. Shunt DC Generator • Field and armature windings are either connected in parallel. 3. Series DC Generator • Field and armature windings are connected in series. 4. Compound DC Generator • Has both shunt and series field so it combines features of series and shunt motors. Equivalent circuit of DC generator Separately excited DC generator IL IA VF IF RF VT EA I A RA Shunt DC generator IL I A IF VT IF RF VT EA I A RA Series DC generator IL IS I A VT E A I A ( RA RS ) Compound DC generator IL I A IF VT IF RF VT EA I A RA Example • A DC shunt generator has shunt field winding resistance of 100Ω. It is supplying a load of 5kW at a voltage of 250V. If its armature resistance is 0.02Ω, calculate the induced e.m.f. of the generator. Solution Given quantities • Terminal voltage, VT = 250V • Field resistance, RF = 100Ω • Armature resistance, RA = 0.22Ω • Power at the load, P = 5kW Solution (cont..) I A IL IF The field current, V T 250V IF 2.5A RF 100 The load current, P 5000 W IL 20A VT 250V The armature current, IA = IL + IF = 20A + 2.5A = 22.5A The induced e.m.f., EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V